x.sf

#!/usr/bin/ruby

func test(n) {
    (powmod(2, n.dec/2, n) == n-1) && n.factor.any {|p|
        #(powmod(2, p.dec/2, p) == p-1) &&
        (p-1 `divides` n.dec/2)
    }
}


__END__

n = p*q

(p-1)*(((p*q - 1)/2) / (p-1)) = (p*q-1)/2


((2*a+1)-1)*((((2*a+1)*(2*b+1) - 1)/2) / ((2*a+1)-1)) = ((2*a+1)*(2*b+1)-1)/2


1/2 ((2 a + 1) (2 b + 1) - 1) = 2 a b + a + b



for k in (1..1e6) {

    #next if k.is_prime

    if (test(k)) {
        #if (k.factor.any {|p| kronecker(2, p) == 1 }) {
        #if (kronecker(2, k) == -1) {

        #if (k.dec `divides` k.dec/2) {

        #if (!k.factor.any {|p|
        #    powmod(2, p.dec/2, p) == p-1
        #}) {
            #p-1 `divides` ((k-1)/2) }) {
            say k
        #}
    }
}

__END__

If 2^((n-1)/2) == -1 (mod n), then p-1 cannot divide (n-1)/2, where p is any prime factor of n.

The criteria for an absolute Euler pseudoprime is that for each prime factor p of n, p-1 divides (n-1)/2.

If n is a an odd composite integer, satisfying:

    2^((n-1)/2) == -1 (mod n)

then, at least one of its prime factors p, must satisfy:

    2^((p-1)/2) == -1 (mod p)

By writing n = p*k, for some k, then:

    p-1 cannot divide p*k-1





It's easy to show that a prime number p, cannot satisfy both conditions:

    if 2^((p-1)/2) == -1 (mod p), then p-1 cannot divide (p-1)/2, since p-1 > (p-1)/2.



If n is an odd composite integer satisfying:

    2^((n-1)/2) == -1 (mod n)

then all its prime factors p also satisfy:

    2^((p-1)/2) == -1 (mod p)


If n is a composite number n = p*q, for two primes p and q, then (p*q - 1)/2 cannot be divisible by p-1.

This can be shown by writing p and q as:

    p = 2*a + 1
    q = 2*b + 1

for some integers a and b.

 (((2*a+1)*(2*b+1) - 1)/2) / (2*a)


If 2^((p*q-1)/2) == -1 (mod p*q), then neither p-1 nor q-1 can divide (p*q-1)/2.

This can be shown by writing p = 2*a + 1, q = 2*a + 1:

    2^(((2*a+1)*(2*b+1)-1)/2) == -1 (mod p*q)

which simplifies to:

    2^(2*a*b + a + b) == -1 (mod p*q)

    2^(2*a*b) * 2^a * 2^b == -1 (mod p*q)

By substituing back a = (p-1)/2 and b = (q-1)/2, we get:

    2^(2*((p-1)/2)*((q-1)/2)) * 2^((p-1)/2) * 2^((q-1)/2) == -1 (mod p*q)





simplifying to:

    2^(1/2 (p - 1) (q - 1) + p/2 + q/2 - 1) == -1 (mod p*q)




#2^((p*q-1)/2) == -1



__END__


Suppose p is a prime p > 2, then p can be expressed as 2*q + 1, for some integer q:

    2^((p-1)/2) == -1 (mod p)

which is equivalent with:

    2^(((2*q + 1) - 1) / 2) == -1 (mod p)

and gets simplified to:

    2^q == -1 (mod p)

Then:
    2^q + 1 == 0 (mod p)


From this we know that:

    2^q + 1 is divisible by p

Therefore:

    2^q + 1 = p * k, for some integer k


    2^((p-1)/2) + 1 = p*k





Suppose p is a prime p > 2, then p can be expressed as 2*q + 1, for some integer q:

    2^(((2*q + 1) - 1) / 2) == -1




__END__

2^((p*q-1)/2) == -1 (mod n)

then:

2^((p*q-1)/2) + 1 = 0 (mod n)




__END__
If n = p*q, then:

    ((p*q - 1)/2) / (p-1) = (p*q - 1) / (2*(p - 1))


Equivalent to:

    2*(p-1) divides p*q-1



__END__

If this were the case, then:

    (n-1)/2 = (p-1)*(q-1)

meaning that:

    n = 2*p*q - 2*p - 2*q + 3




Yes, there doesn't seem to exist any numbers n such that satisfy both conditions:

    2^((n-1)/2) == -1 (mod n)

and

    (p-1) divides (n-1)/2 for all prime factors p of n

I think this is easy to prove:



var primes_bellow = []

func isok(k) {
    var w = powmod(prime(k), (k-1)/2, k)

    (w != 1) && (w != k-1) || return false

    primes_bellow.all {|p|
        var t = powmod(p, (k-1)/2, k)
        (t == 1) || (t == k-1)
    }
}

var p = 2;

for k in (1..1e7) {
    if (isok(k)) {
        say k

        primes_bellow << p
        p = p.next_prime
    }
}

# (($n-1)>>1) % ($_-1) == 0





However, by slightly weaking the second condition:

    (p-1) dividies (n-1) for all prime factors p of n





...then the first few terms would be:
    29341, 314821, 15247621, 36765901, 133800661, 299736181, 579606301, 702683101, 739444021, 775368901, 4127050621, 4340265931, 4953963781, 7361854501, 9293756581, 11360308765, 13734086221, 14386156093, 18173966581, 19331388805, 20193897781, 23224518901, 24800417461, 27363972901, 34087042891, 37870128451, 41135881645, 41764915501, 42304192981, 46493311411, 48354810571, 52396612381, 53038112821, 63734663077, 65376249445, 72725349421, 74500799437, 90495831037, 94706509885, 96133341781, 117930513949, 118805562613, 118936706941, 245291853691, 268628041501, 347556640861, 366552764101, 370230237469, 396124060501, 436580025661, 440688939301, 443729109421, 494442433171, 547365250405, 600613114501, 638345536501, 705636161413, 707161856941, 716726903707, 779999961061, 813460927501, 1084962050965, 1142877413581, 1341908088445, 1427823735229, 1435738307221, 1479982594861, 1557309989845, 1557325889821, 1671885346141, 1683061731541, 1766297137501, 1918081357501, 2030606255101, 2100917530531, 2152302898747, 2289717532901, 2332465835701, 2432870015941, 2657407435621, 2874412113301, 2881383743701, 2951861939461, 3305918887501, 3459443867461, 3944036145061, 4018209994261, 4038030605701, 4443982738651, 4525783922251, 4963134761131, 5069712129301, 5459748937021, 5576047603741, 5811733744867, 6317414663701, 6513448976101, 6614815335661, 7835972164021, 8333344148677